Nucleic Acid Biochemistry — MCQs

Nucleic Acid Biochemistry Indian Medical PG Practice Questions and MCQs

Question 181: What enzyme deficiency is associated with the overproduction of uric acid and the development of gout?

A. Adenosine deaminase
B. Hypoxanthine-guanine phosphoribosyltransferase (Correct Answer)
C. Xanthine oxidase
D. Uricase

Explanation: ***Hypoxanthine-guanine phosphoribosyltransferase (HGPRT)*** - A deficiency in **HGPRT** leads to an inability to salvage hypoxanthine and guanine, shunting these purine bases towards degradation. - This results in the **overproduction of uric acid** as the end product of purine metabolism, directly contributing to hyperuricemia and gout. *Adenosine deaminase* - Deficiency of **adenosine deaminase (ADA)** leads to the accumulation of deoxyadenosine and its derivatives, which are toxic to lymphocytes. - This causes **severe combined immunodeficiency (SCID)**, not gout or uric acid overproduction. *Xanthine oxidase* - **Xanthine oxidase** is the enzyme responsible for the final steps of purine degradation, converting hypoxanthine to xanthine and then xanthine to uric acid. - Inhibition of xanthine oxidase (e.g., by allopurinol) is a treatment for gout, meaning a **deficiency would *reduce* uric acid production**, not increase it. *Uricase* - **Uricase** is an enzyme that converts uric acid into allantoin, a more soluble compound. - Humans naturally lack this enzyme, which is why we excrete uric acid; its absence is normal and not a "deficiency" causing *overproduction* of uric acid, although some conditions leverage recombinant uricase for treatment.

Question 182: Which of the following statements regarding mitochondrial DNA is FALSE?

A. Double stranded
B. Inherited from mother
C. High mutation rate
D. All respiratory proteins are synthesized within the mitochondria (Correct Answer)

Explanation: ***All respiratory proteins are synthesized within the mitochondria.*** - While mitochondrial DNA (mtDNA) encodes some proteins essential for the **electron transport chain** (respiratory proteins), not all respiratory proteins are synthesized within the mitochondria. - Many crucial respiratory proteins are encoded by **nuclear DNA** and imported into the mitochondria from the cytoplasm. *Double stranded* - **Mitochondrial DNA (mtDNA)** is a **double-stranded**, circular molecule, similar to bacterial chromosomes. - This structure provides stability and allows for efficient replication within the organelle. *Inherited from mother* - Mitochondria and their DNA are exclusively inherited from the **mother** during fertilization, as sperm primarily contributes nuclear DNA. - This **maternal inheritance pattern** is a key feature of mtDNA and is used in tracing ancestry. *High mutation rate* - mtDNA has a significantly **higher mutation rate** compared to nuclear DNA due to several factors, including lack of robust repair mechanisms and exposure to reactive oxygen species. - This contributes to the rapid evolution of mtDNA and its use in **population genetics** studies.

Question 183: Which of the following is NOT a function of DNA polymerase I?

A. Involved in the synthesis of Okazaki fragments.
B. Repairs damaged DNA.
C. Participates in DNA replication.
D. Not essential for bacterial survival. (Correct Answer)

Explanation: ***Not essential for bacterial survival.*** - While DNA polymerase I is important, it is **not absolutely essential** for _E. coli_ survival because **DNA polymerase III** carries out the bulk of replication, and other repair enzymes can compensate for some of its repair functions. - "Not essential for bacterial survival" is a **characteristic/property**, not a **function** of the enzyme, making it the correct answer to this "NOT a function" question. - However, its absence does lead to a **mutator phenotype** and increased sensitivity to DNA damaging agents. *Involved in the synthesis of Okazaki fragments.* - DNA polymerase I **IS** involved in **Okazaki fragment processing** on the lagging strand. - It removes RNA primers using its **5' → 3' exonuclease** activity and fills the resulting gaps with DNA nucleotides. - The initial synthesis (elongation) of Okazaki fragments is carried out by **DNA polymerase III**, but Pol I completes their maturation. *Repairs damaged DNA.* - DNA polymerase I possesses both **5' → 3' exonuclease** and **3' → 5' exonuclease** activity, allowing it to remove damaged bases and incorporate new DNA. - It plays a crucial role in various DNA repair mechanisms, including **nucleotide excision repair** and **base excision repair**. *Participates in DNA replication.* - DNA polymerase I is essential for DNA replication, specifically in the **maturation of Okazaki fragments** on the lagging strand by removing RNA primers and filling gaps. - It also contributes to the **proofreading** and repair of newly synthesized DNA during replication.

Question 184: In which of the following cellular components is phosphorus present?

A. RNA
B. Cell membrane
C. DNA
D. All of the options (Correct Answer)

Explanation: ***All of the options*** - Phosphorus is an essential element present in **multiple cellular components**, making this the correct comprehensive answer. - It is a key component of **phospholipids** (cell membrane), **DNA backbone**, and **RNA backbone**. *Why not just "Cell membrane"?* - The cell membrane contains phosphorus in its **phospholipid bilayer** (phosphatidylcholine, phosphatidylethanolamine, etc.). - While correct that phosphorus is present here, it's **incomplete** as an answer since phosphorus is also in DNA and RNA. *Why not just "DNA"?* - DNA contains phosphorus in the **phosphate groups** that link deoxyribose sugars in the sugar-phosphate backbone. - This is medically accurate but **incomplete** since phosphorus is also present in RNA and cell membranes. *Why not just "RNA"?* - RNA contains phosphorus in the **phosphate groups** that link ribose sugars in its backbone structure. - While true, this is **incomplete** as phosphorus is also essential in DNA and membrane phospholipids.

Question 185: Which of the following statements about purine synthesis is true?

A. Glutamine PRPP amidotransferase is the rate-limiting enzyme in purine synthesis.
B. THFA derivatives are required for the formation of C2 and C8 in the purine ring.
C. Glutamine donates the amino group for N1 of the purine ring.
D. IMP is the first nucleotide synthesized during purine synthesis. (Correct Answer)

Explanation: ***IMP is the first nucleotide synthesized during purine synthesis.*** - **Inosine monophosphate (IMP)** is the first complete purine nucleotide formed in de novo purine synthesis. - It serves as the precursor for both **AMP** and **GMP**, making it the foundational molecule in the purine biosynthesis pathway. - The synthesis pathway converges at IMP before branching to form the specific adenine and guanine nucleotides. *Glutamine PRPP amidotransferase is the rate-limiting enzyme in purine synthesis.* - While **glutamine PRPP amidotransferase** catalyzes the committed step in de novo purine synthesis, it is often considered a key regulatory enzyme. - However, the statement is marked incorrect in this context because **PRPP synthetase** (which forms PRPP from ribose-5-phosphate) can also be considered rate-limiting depending on PRPP availability. - The first committed step specific to purines is the glutamine PRPP amidotransferase reaction, making this a debatable but commonly accepted alternative answer. *THFA derivatives are required for the formation of C2 and C8 in the purine ring.* - **Tetrahydrofolate (THF) derivatives** do provide one-carbon units to positions **C2 and C8** of the purine ring. - Specifically, **N10-formyl-THF** donates the formyl group for C2, and **N5,N10-methenyl-THF** (which converts to N10-formyl-THF) provides C8. - This statement is technically correct, but may be marked incorrect if the question seeks a more fundamental defining feature of purine synthesis (such as IMP being the first nucleotide). *Glutamine donates the amino group for N1 of the purine ring.* - **Glutamine** provides nitrogen atoms at positions **N3 and N9** of the purine ring. - The **N1 nitrogen** is derived from the amino group of **aspartate**, not glutamine. - This statement is clearly incorrect.

Question 186: What type of linkage connects individual nucleotides in a polynucleotide chain?

A. 5'-3' Phosphodiester linkage
B. 3'-5' Phosphodiester linkage (Correct Answer)
C. N-glycosidic linkage
D. N-glycosidic bond

Explanation: ***3'-5' Phosphodiester linkage*** - This is the correct linkage that connects individual nucleotides in a polynucleotide chain. - The phosphodiester bond forms between the **3'-hydroxyl (OH) group** of the pentose sugar of one nucleotide and the **5'-phosphate group** of the adjacent nucleotide. - This creates the **sugar-phosphate backbone** of DNA and RNA, providing structural integrity and directionality to the polynucleotide chain. - The linkage is named **3'-5'** because it connects the **3' carbon** of one sugar to the **5' carbon** of the next sugar via a phosphate group. *5'-3' Phosphodiester linkage* - This term is **misleading** when describing the linkage itself. - While DNA synthesis and chain reading occur in the **5' to 3' direction**, the actual chemical bond is a **3'-5' phosphodiester linkage**. - The 5'-3' notation refers to the **direction of chain growth**, not the name of the linkage connecting nucleotides. *N-glycosidic linkage* - This bond connects the **nitrogenous base** to the **1' carbon** of the pentose sugar (ribose or deoxyribose). - It is essential for forming a nucleoside and nucleotide, but does **not** link one nucleotide to another in the polynucleotide chain. - This is an intramolecular bond within a single nucleotide, not an internucleotide linkage. *N-glycosidic bond* - This is the same as N-glycosidic linkage—it connects the **base to the sugar** within a single nucleotide. - It does not connect adjacent nucleotides together in a polynucleotide chain. - The bond is between the N9 of purines or N1 of pyrimidines and the C1' of the sugar.

Question 187: What is the primary functional mechanism of small RNAs in cellular processes?

A. Typically less than 200 nucleotides in length
B. Involved in post-transcriptional regulation of gene expression
C. Always synthesized endogenously
D. Regulate gene expression through RNA interference mechanisms (Correct Answer)

Explanation: ***Regulate gene expression through RNA interference mechanisms*** - Small RNAs, such as **miRNAs (microRNAs)** and **siRNAs (small interfering RNAs)**, primarily function by guiding **RNA-induced silencing complexes (RISC)** to target mRNA molecules. - This interaction leads to either the **degradation of mRNA** or the **inhibition of its translation**, thereby regulating gene expression at the post-transcriptional level. *Typically less than 200 nucleotides in length* - While small RNAs are indeed short, this statement describes a **physical characteristic** rather than their primary cellular function. - Their function is linked to specific molecular interactions, not simply their size. *Involved in post-transcriptional regulation of gene expression* - This is a correct statement about their function but is **less specific** than regulating gene expression via RNA interference. - RNA interference is the specific mechanism by which many small RNAs achieve post-transcriptional regulation. *Always synthesized endogenously* - While many small RNAs like miRNAs are **endogenously synthesized**, some siRNAs can be **exogenously introduced** (e.g., in research) or derived from viral infections. - Therefore, stating they are *always* synthesized endogenously is inaccurate.

Question 188: Which of the following statements about pyrimidine catabolism is correct?

A. It directly contributes to carnosine synthesis.
B. β-aminoisobutyrate is produced during the process. (Correct Answer)
C. It generates intermediates of the citric acid cycle.
D. It is the primary source of uric acid production.

Explanation: ***β-aminoisobutyrate is produced during the process.*** - **β-aminoisobutyrate** is a direct and prominent end-product of **thymine catabolism**, which is a pyrimidine base. - This compound is specifically characteristic of pyrimidine degradation and is not produced by other major metabolic pathways. - Its presence in urine can be an indicator of increased DNA turnover or a genetic deficiency in **dihydropyrimidine dehydrogenase**. - This is the most specific and direct statement about pyrimidine catabolism among the options. *It generates intermediates of the citric acid cycle.* - While pyrimidine catabolism products (**β-alanine** and **β-aminoisobutyrate**) can be further metabolized through multiple enzymatic steps to eventually form **succinyl-CoA** (a TCA cycle intermediate), this is an **indirect** process requiring transamination and CoA activation. - This is not a primary or direct feature of pyrimidine catabolism itself, but rather a downstream metabolic fate of its products. - In contrast, **β-aminoisobutyrate production** is a direct, immediate result of pyrimidine breakdown, making option A more specifically correct. *It directly contributes to carnosine synthesis.* - **Carnosine** is a dipeptide composed of **β-alanine** (a product of cytosine/uracil catabolism) and **histidine**. - While **β-alanine** from pyrimidine catabolism can be used for carnosine synthesis, the word **"directly"** makes this statement incorrect. - β-alanine must first be activated and undergo peptide bond formation with histidine through carnosine synthase, making this an indirect contribution. *It is the primary source of uric acid production.* - **Uric acid** is the end-product of **purine catabolism**, not pyrimidine catabolism. - Pyrimidine catabolism results in completely different end-products: **β-alanine**, **β-aminoisobutyrate**, **CO2**, and **NH3** (ammonia). - This statement confuses the two distinct nucleotide degradation pathways.

Question 189: Which type of DNA is characterized by a left-handed helix?

A. B-DNA
B. A-DNA
C. Z-DNA (Correct Answer)
D. F-DNA

Explanation: ***Z-DNA*** - **Z-DNA** is a unique form of DNA characterized by its distinctive **left-handed helical structure**, as opposed to the right-handed helices of A-DNA and B-DNA. - This structure forms under specific conditions, often in regions with alternating **purine-pyrimidine sequences** (e.g., GCGCGC) and is thought to play roles in **gene regulation** and chromatin structure. - It has a **zigzag backbone** appearance, which gives it its name. *B-DNA* - **B-DNA** is the most common and **biologically significant form of DNA** found under physiological conditions in living cells. - It exhibits a **right-handed helical structure**, with approximately 10-10.5 base pairs per turn. - This is the predominant form in aqueous environments at neutral pH. *A-DNA* - **A-DNA** is a **right-handed helical DNA** form that is usually found under dehydrating conditions or when DNA is complexed with certain proteins. - It has a wider and shorter helix than B-DNA, with about 11 base pairs per turn. - The bases are tilted relative to the helix axis. *F-DNA* - **F-DNA** is not a recognized standard DNA conformation in biochemistry. - The three well-established DNA forms are **A-DNA, B-DNA, and Z-DNA**, based on their structural characteristics and biological relevance.

Question 190: Bond formation between ribose sugar and nitrogen is ?

A. Phosphodiester linkage
B. Phosphoester linkage
C. Glycosidic linkage (Correct Answer)
D. Acid anhydride linkage

Explanation: ***Glycosidic linkage*** - A **glycosidic bond** forms between the **anomeric carbon** of a carbohydrate (like **ribose**) and another functional group (like the **nitrogenous base**). - Specifically, this bond links the **1' carbon atom** of the ribose sugar to a **nitrogen atom** (N-1 in pyrimidines, N-9 in purines) of the nitrogenous base to form a **nucleoside**. *Phosphodiester linkage* - This bond connects the **5' carbon** of one sugar to the **3' carbon** of another sugar via a **phosphate group**, forming the backbone of **DNA and RNA**. - It involves a **phosphate** and **two ester bonds**, not directly linking sugar to a nitrogenous base. *Phosphoester linkage* - A **phosphoester bond** is formed when a **phosphate group** reacts with a **hydroxyl group** of an alcohol, typically found in molecules like **nucleotides** (e.g., 5'-phosphate of a nucleoside). - This type of bond is part of the **phosphodiester linkage** but does not describe the bond between the sugar and the nitrogenous base. *Acidanhydride linkage* - An **acid anhydride linkage** is formed between **two acid groups** with the elimination of water, such as in **ATP** where two phosphate groups are linked by this high-energy bond. - This type of bond is not involved in the connection between a **ribose sugar** and a **nitrogenous base**.

Practice by Chapter

Nucleotide Structure and Function

Practice Questions

DNA Structure and Replication

Practice Questions

RNA Structure and Types

Practice Questions

Transcription: RNA Synthesis

Practice Questions

Post-Transcriptional Modifications

Practice Questions

Translation: Protein Synthesis

Practice Questions

Genetic Code and Codon Usage

Practice Questions

Regulation of Gene Expression

Practice Questions

Mutations and DNA Repair

Practice Questions

Purine Metabolism and Disorders

Practice Questions

Pyrimidine Metabolism and Disorders

Practice Questions

Nucleotide Degradation and Salvage Pathways

Practice Questions

Want unlimited practice?

Get full access to all questions, explanations, and performance tracking.

Start For Free